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The sine inverse of a purely complex number

suvadip

Member
Feb 21, 2013
69
To prove that
\(\displaystyle sin^{-1}(ix)=2n\pi\pm i log(\sqrt{1+x^2}+x)\)
I can prove \(\displaystyle sin^{-1}(ix)=2n\pi+ i log(\sqrt{1+x^2}+x)\)

but facing problem to prove
\(\displaystyle sin^{-1}(ix)=2n\pi- i log(\sqrt{1+x^2}+x)\)

Help please
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: Complex no

To prove that
\(\displaystyle sin^{-1}(ix)=2n\pi\pm i log(\sqrt{1+x^2}+x)\)
I can prove \(\displaystyle sin^{-1}(ix)=2n\pi+ i log(\sqrt{1+x^2}+x)\)

but facing problem to prove
\(\displaystyle sin^{-1}(ix)=2n\pi- i log(\sqrt{1+x^2}+x)\)

Help please
Can you sketch your proof ? what is $n$ ?
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Hint : $e^{ix} = \cos(x) + i\sin(x)$.

ZaidAlyafey said:
what is n?
An arbitrary integer of course.
 

suvadip

Member
Feb 21, 2013
69
Re: Complex no

Can you sketch your proof ? what is $n$ ?
n is any integer

Let \(\displaystyle sin^{-1}(ix)=\theta\)

Then \(\displaystyle sin\theta = ix\)

\(\displaystyle cos\theta=\pm\sqrt{1-sin^2\theta}=\pm\sqrt{1+x^2}\)

Now \(\displaystyle e^{i\theta}=cos\theta +i sin\theta=\pm\sqrt{1+x^2}+i i x=-x\pm\sqrt{1+x^2}\)

Therefore \(\displaystyle i\theta=Log (-x+\sqrt{1+x^2})(Only + sign ~~considered)=2n\pi i +log(-x+\sqrt{1+x^2})\)

\(\displaystyle \theta = 2n\pi-i log(-x+\sqrt{1+x^2})\)
\(\displaystyle \theta = 2n\pi+i log(x+\sqrt{1+x^2})\)
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
In order to get the other sign, simply note that $\sin(-x) = -\sin(x)$.
 

suvadip

Member
Feb 21, 2013
69